SEVERAL DEPENDENT VARIABLES; LAGRANGE’S EQUATIONS

5. SEVERAL DEPENDENT VARIABLES; LAGRANGE’S EQUATIONS

It is not necessary to restrict ourselves to problems with one dependent variable y. Recall that in ordinary calculus problems the necessary condition for a minimum point on z = z(x) is dz/dx = 0; for a function of two variables z = z(x, y), we have the two conditions ∂z/∂x = 0 and ∂z/∂y = 0. We have a somewhat analogous situation in the calculus of variations. Suppose that we are given an F which is a function of y, z, dy/dx, dz/dx, and x, and we want to find two curves y = y(x) and z = z(x) which make I = depends on both y(x) and z(x) and you might very well guess that in this case we would have two Euler equations, one for y and one for z, namely

By carrying through calculations similar to those we used in deriving the single Euler equation for the one dependent variable case you can show (Problem 1(a)) that this guess is correct. If there are still more dependent variables (but one independent variable), then we write an Euler equation for each dependent variable. It is also possible to consider a problem with more than one independent variable (see Problem 1(b)) or with F depending on y ′′ as well as x, y, y ′ (see Problem 1(c)).

There is a very important application of equations like (5.1) to mechanics. In elementary physics, Newton’s second law F = ma is a fundamental equation. In more advanced mechanics, it is often useful to start from a different assumption (which can be proved equivalent to Newton’s law; see mechanics text books.) This assumption is called Hamilton’s principle. It says that any particle or system of

t particles always moves in such a way that I = 2 t 1 L dt is stationary, where L = T −V is called the Lagrangian; T is the kinetic energy, and V is the potential energy of

the particle or system. Example 1. Use Hamilton’s principle to find the equations of motion of a single particle

of mass m moving (near the earth) under gravity. We first write the formulas for the kinetic energy T and the potential energy V of the particle. (It is convenient to use a dot to mean a derivative with respect to t just as we use a prime to indicate a derivative with respect to x; thus dx/dt = ˙x,

dy/dt = ˙y, dy/dx = y ′ ,d 2 x/dt 2 =¨ x, etc.) The equations for T , V , and L = T − V , are:

T= mv 2 = 1 m( ˙x 2 + ˙y 2 + ˙z 2 ),

V = mgz,

L=T−V= m( ˙x 2 + ˙y 2 + ˙z 2 2 ) − mgz. Here t is the independent variable; x, y, and z are the dependent variables, and

t L corresponds to what we have called F previously. Then to make I = 2

t 1 L dt stationary, we write the corresponding Euler equations. There are three Euler

equations, one for x, one for y, and one for z. The Euler equations are called Lagrange’s equations in mechanics [see (5.3) next page].

486 Calculus of Variations Chapter 9

Lagrange’s equations

Substituting L in (5.2) into Lagrange’s equations (5.3), we get

(m ˙x) = 0

  d ˙x = const., (5.4)

˙y = const.,

These are just the familiar equations obtained from Newton’s law; they say that in the gravitational field near the surface of the earth, the horizontal velocity is constant and the vertical acceleration is −g. In this problem you may say that it would have been simpler just to write the equations from Newton’s law in the first place! This is true in simple cases, but in more complicated problems it may be much simpler to find one scalar function (that is, L) than to find six functions (that is, the components of the two vectors, force and acceleration). For example, the acceleration components in spherical coordinates are quite complicated to derive by elementary methods (see mechanics text books), but you should have no trouble deriving the equations of motion in polar, cylindrical or spherical coordinates using the Lagrangian. Let’s do some examples.

Example 2. Use Lagrange’s equations to find the equations of motion of a particle in terms of the polar coordinate variables r and θ. The element of arc length in polar coordinates is ds where

ds 2 = dr 2 +r 2 dθ 2 .

The velocity of a moving particle is ds/dt; from (5.5) we get

The kinetic energy is 1 2 mv 2 , so we have

T= m( ˙r 2 +r 2 ˙θ 2 ),

L=T−V= m( ˙r 2 +r 2 ˙θ 2 2 ) − V (r, θ),

Section 5 Several Dependent Variables; Lagrange’s Equations 487

where V (r, θ) is the potential energy of the particle. Lagrange’s equations in the variables r, θ are:

Substituting L from (5.7) into (5.8), we get

The r equation of motion is, then,

2 (5.10) ∂V m(¨

r − r ˙θ )=−

∂r

The θ equation is

m(r 2 θ + 2r ˙r ˙θ) = − ¨ ∂V , ∂θ

or, dividing by r,

Now the quantities −∂V/∂r and −(1/r)(∂V/∂θ) are the components of the force (F = −∇V ) on the particle in the r and θ directions. (See Chapter 6.) Then equations (5.10) and (5.11) are just the components of ma = F; the acceleration components are then

The second term in a r is a familiar one; it is just the centripetal acceleration v 2 /r when v = r ˙θ (the minus sign indicates that it is toward the origin). The second term in a θ is called the Coriolis acceleration.

We show by an example another important point about Lagrange’s equations.

Example 3.

A mass m 1 moves without friction on the surface

of the cone shown (Figure 5.1). Mass m 2 is joined to m 1 by a string of constant length; m 2 can move only

vertically up and down. Find the Lagrange equations of motion of the system.

Let’s use spherical coordinates ρ, θ, φ for m 1 , and coordinate z for m 2 . Then for m 1 ,v 2 = (ds/dt) 2 = ˙ρ 2 +ρ 2 ˙θ 2 2 +ρ 2 sin θ˙ φ 2 [Chapter 5, equation (4.20)], and for m 2 ,v 2 = ˙z 2 . The potential energy mgh of m 1 is

m 1 gρ cos θ and of m 2 is m 2 gz. Note that we have used

Figure 5.1

488 Calculus of Variations Chapter 9

four variables: ρ, θ, φ, z; however, there are not four Lagrange equations We must use the equation of the cone (θ = 30 ◦ ) and the equation ρ + |z| = l (string of constant length) to eliminate θ and either ρ or z. The Lagrangian L must always

be written using the smallest possible number of variables (we say that we eliminate the constraint equations). Then, with θ = 30 ◦ , sin θ = 1 , cos θ = 1

2 2 3, ˙θ = 0, and z = −|z| = −(l − ρ), we find L in terms of ρ and φ:

L= m 1 ( ˙ρ +ρ φ /4) + m 2 ˙ρ − m 1 gρ 3+m 2 g(l − ρ).

Thus the Lagrange equations are

(m 1 ˙ρ + m 2 ˙ρ) − m 1 ρ˙ φ /4 + m 1 g 3+m 2 g = 0, dt

(m ρ 2 1 φ/4) = 0 ˙

or ρ 2 φ = const. ˙

dt

PROBLEMS, SECTION 5

1. (a) Consider the case of two dependent variables. Show that if F = F (x, y, z, y ′ ′ R x ,z ) and we want to find y(x) and z(x) to make I =

1 F dx stationary, then y and z should each satisfy an Euler equation as in (5.1). Hint: Construct a formula

for a varied path Y for y as in Section 2 [Y = y + ǫη(x) with η(x) arbitrary] and construct a similar formula for z [let Z = z + ǫζ(x), where ζ(x) is another arbitrary function]. Carry through the details of differentiating with respect to ǫ, putting ǫ = 0, and integrating by parts as in Section 2; then use the fact that both η(x) and ζ(x) are arbitrary to get (5.1).

(b) Consider the case of two independent variables. You want to find the function u(x, y) which makes stationary the double integral

F (u, x, y, u x ,u y ) dx dy.

Hint: Let the varied U (x, y) = u(x, y) + ǫη(x, y) where η(x, y) = 0 at x = x 1 , x=x 2 ,y=y 1 ,y=y 2 , but is otherwise arbitrary. As in Section 2, differentiate with respect to ǫ, set ǫ = 0, integrate by parts, and use the fact that η is arbitrary. Show that the Euler equation is then

(c) Consider the case in which F depends on x, y, y ′ , and y ′′ . Assuming zero values of the variation η(x) and its derivative at the endpoints x 1 and x 2 , show that then the Euler equation becomes

2. Set up Lagrange’s equations in cylindrical coordinates for a particle of mass m in a potential field V (r, θ, z). Hint: v = ds/dt; write ds in cylindrical coordinates.

3. Do Problem 2 in spherical coordinates. 4. Use Lagrange’s equations to find the equation of motion of a simple pendulum. (See

Chapter 7, Problem 2.13.)

Section 5 Several Dependent Variables; Lagrange’s Equations 489

5. Find the equation of motion of a particle moving along the x axis if the potential

energy is V = 1 2 kx 2 . (This is a simple harmonic oscillator.)

6. A particle moves on the surface of a sphere of radius a under the action of the earth’s gravitational field. Find the θ, φ equations of motion. (Comment: This is called a spherical pendulum. It is like a simple pendulum suspended from the center of the sphere, except that the motion is not restricted to a plane.)

7. Prove that a particle constrained to stay on a surface f (x, y, z) = 0, but subject to no other forces, moves along a geodesic of the surface. Hint: The potential energy V is constant, since constraint forces are normal to the surface and so do no work on the particle. Use Hamilton’s principle and show that the problem of finding a geodesic and the problem of finding the path of the particle are identical mathematics problems.

8. Two particles each of mass m are connected by an (in- extensible) string of length l. One particle moves on a horizontal table (assume no friction), The string passes through a hole in the table and the particle at the lower end moves up and down along a vertical line. Find the Lagrange equations of motion of the particles. Hint: Let the coordinates of the particle on the table be r and θ, and let the coordinate of the other particle be z. Eliminate one variable from L (using r + |z| = l) and write two Lagrange equations.

9. A mass m moves without friction on the surface of the cone r = z under gravity acting in the negative z direction. Here r is the cylindrical coordinate r = px 2 +y 2 . Find the Lagrangian and Lagrange’s equations in terms of r and θ (that is, elimi- nate z).

10. Do Example 3 above, using cylindrical coordinates for m 1 . Hint: Use z 1 and z 2 for the z coordinates of m 1 and m 2 . What is the equation of the cone in terms of r and z 1 coordinates (see Chapter 5, Figures 4.4 and 4.5).

11. A yo-yo (as shown) falls under gravity. Assume that it falls straight down, unwinding as it goes. Find the Lagrange equation of motion. Hints: The kinetic energy is the sum of the trans-

lational energy 1 2 m ˙z 2 and the rotational energy 1 2 2 where I is I ˙θ the moment of inertia. What is the relation between ˙z and ˙θ? Assume the yo-yo is a solid cylinder with inner radius a and outer radius b.

12. Find the Lagrangian and Lagrange’s equations for a simple pendulum (Problem 4) if the cord is replaced by a spring with spring constant k. Hint: If the unstretched

spring length is r 0 , and the polar coordinates of the mass m are (r, θ), the potential

energy of the spring is 1 2 k(r − r 0 ) 2 .

13. A particle moves without friction under gravity on the surface of the paraboloid z=x 2 +y 2 . Find the Lagrangian and the Lagrange equations of motion. Show that motion in a horizontal circle is possible and find the angular velocity of this motion. Use cylindrical coordinates.

490 Calculus of Variations Chapter 9

14. A hoop of mass M and radius a rolls without slipping down an inclined plane of angle α. Find the Lagrangian and the Lagrange equation of motion. Hint: The kinetic energy of a body which is both translating and rotating is a sum of two

terms: the translational kinetic energy 1 2 Mv 2 where v is the velocity of the center of mass, and the rotational kinetic energy 1 2 Iω 2 where ω is the angular velocity and I is the moment of inertia around the rotation axis through the center of mass.

15. Generalize Problem 14 to any mass M of circular cross section and moment of inertia I. Consider a hoop, a disk, a spherical shell, a solid spherical ball; order them as to which would first reach the bottom of the inclined plane. (For moments of inertia, see Chapter 5, Section 4.)

16. a Find the Lagrangian and the Lagrange equation for ␪ the pendulum shown. The vertical circle is fixed. The

string winds up or unwinds as the mass m swings back and forth. Assume that the unwound part of the string at any time is in a straight line tangent

lm to the circle. Let l be the length of unwound string when the pendulum hangs straight down.

m 17. A simple pendulum (Problem 4) is suspended from a mass M which is free to move

without friction along the x axis. The pendulum swings in the xz plane and gravity acts in the negative z direction. Find the Lagrangian and Lagrange’s equations for the system.

18. A hoop of mass m in a vertical plane rests on a frictionless table. A thread is wound many times around the circumference of the hoop. The free end of the thread extends from the bottom of the hoop along the table, passes over a pulley (assumed weightless), and then hangs straight down with a mass m (equal to the mass of the hoop) attached to the end of the thread. Let x be the length of thread between the bottom of the hoop and the pulley, let y be the length of thread between the pulley and the hanging mass m, and let θ be the angle of rotation of the hoop about its center if the thread unwinds. What is the relation between x, y, and θ? Find the Lagrangian and Lagrange’s equations for the system. If the system starts from rest, how does the hoop move?

For the following problems, use the Lagrangian to find the equations of motion and then refer to Chapter 3, Section 12.

19. For small vibrations, find the characteristic frequencies and the characteristic modes of vibration of the coupled pendulums shown. All motion takes place in a sin- gle vertical plane. Assume the spring unstretched when both pendulums hang vertically, and take the spring constant as k = mg/l to simplify the algebra. Hints: Write the kinetic and potential energies in terms of the rectangular coordinates of the masses relative to their positions hanging at rest. Don’t forget the gravitational potential energies. Then write the rectangular coordinates x and y in terms of θ and φ, and for small vibra-

tions approximate sin θ = θ, cos θ = 1 − θ 2 /2, and similar equations for φ.

Section 6 Isoperimetric Problems 491

20. Do Problem 19 if the spring constant is k = 3mg/l. 21. Find the Lagrangian and Lagrange’s equations for

the double pendulum shown. All motion takes ␪ l

place in a single vertical plane. Hint: See the hint in Problem 19. m

22. Do Problem 21 if the two masses are different. Let ␾ l

m be the lower mass and let M be the sum of the m two masses.

23. For small oscillations of the double pendulum in Problem 22, let M = 4m and find the characteristic frequencies and characteristic modes of vibration.

24. Do Problem 23 if M/m = 9/4 25. Do Problem 23 in general, that is, in terms of the ratio M/m. Hint: You may find

it helpful to use a single letter to represent pm/M, say α 2 = m/M .